The integral int frac{2x^3 - 1}{x^4 + x} ,dx | vedclass

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Question

(C) Let $I = \int \frac{2x^3 - 1}{x^4 + x} \,dx$. Divide the numerator and denominator by $x^2$: $I = \int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}

Vedclass · Accepted Answer

$\log_e \left| \frac{x^3 + 1}{x} \right| + C$

Vedclass · Answer

(C) Let $I = \int \frac{2x^3 - 1}{x^4 + x} \,dx$. Divide the numerator and denominator by $x^2$: $I = \int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} \,dx$. Let $t = x^2 + \frac{1}{x}$. Then $dt = (2x - \frac{1}{x^2}) \,dx$. Substituting these into the integral,we get: $I = \int \frac{1}{t} \,dt = \log_e |t| + C$. Substituting back $t = x^2 + \frac{1}{x} = \frac{x^3 + 1}{x}$: $I = \log_e \left| \frac{x^3 + 1}{x} \right| + C$.

The integral $\int \frac{2x^3 - 1}{x^4 + x} \,dx$ is equal to (Here $C$ is a constant of integration)

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